Hippopedal intensity plots: drawing comparisons between antenna and optical polarimetry

Muhammad Hamza Waseem; Faizan-e- Ilahi; Muhammad Sabieh Anwar

doi:10.1364/AO.58.008442

Applied Optics
Vol. 58,
Issue 31,
pp. 8442-8448
(2019)
•https://doi.org/10.1364/AO.58.008442

Hippopedal intensity plots: drawing comparisons between antenna and optical polarimetry

Muhammad Hamza Waseem, Faizan-e- Ilahi, and Muhammad Sabieh Anwar

Not Accessible

Your library or personal account may give you access

Get PDF
Email
Share
Get Citation
Copy Citation Text
Muhammad Hamza Waseem, Faizan-e- Ilahi, and Muhammad Sabieh Anwar, "Hippopedal intensity plots: drawing comparisons between antenna and optical polarimetry," Appl. Opt. 58, 8442-8448 (2019)

Export Citation
- BibTex
- Endnote (RIS)
- HTML
- Plain Text
Citation alert
Save article

Check for updates

Abstract

We present a graphical method for determining the polarization state of light. We review polarization profiling techniques used for antenna systems and correlate them with an optical analog. The analogy is used to perform tabletop optical polarimetry experiments. The experiments require intensity measurements and generation of hippopedal polar plots. The measured polarimetry results are found to agree well with the theoretical predictions. We also set up a cost-effective quantum version of the experiment and reformulate the problem as that of quantum state measurement. These experiments can be incorporated in the teaching laboratory for both classical optics and quantum optics courses, offering a purely quantitative, albeit graphic, technique for polarization analysis. These techniques are already employed for design of microwave antennas; the optic analogy is, therefore, interesting and useful.

Full Article | PDF Article

More Like This

Analysis of errors in polarimetry using a rotating waveplate

Yu Liang, Zhongquan Qu, Yue Zhong, Zhiming Song, and Shaoying Li
Appl. Opt. 58(36) 9883-9895 (2019)

Review of passive imaging polarimetry for remote sensing applications

J. Scott Tyo, Dennis L. Goldstein, David B. Chenault, and Joseph A. Shaw
Appl. Opt. 45(22) 5453-5469 (2006)

Stokes imaging polarimetry using a twisted hybrid aligned nematic liquid crystal cell

Michinori Honma, Yuta Kasai, and Toshiaki Nose
Appl. Opt. 57(32) 9649-9656 (2018)

Previous Article Next Article

Cited By

You do not have subscription access to this journal. Cited by links are available to subscribers only. You may subscribe either as an Optica member, or as an authorized user of your institution.

Contact your librarian or system administrator
or
Login to access Optica Member Subscription

Figures (6)

You do not have subscription access to this journal. Figure files are available to subscribers only. You may subscribe either as an Optica member, or as an authorized user of your institution.

Contact your librarian or system administrator
or
Login to access Optica Member Subscription

Tables (2)

You do not have subscription access to this journal. Article tables are available to subscribers only. You may subscribe either as an Optica member, or as an authorized user of your institution.

Contact your librarian or system administrator
or
Login to access Optica Member Subscription

Equations (29)

You do not have subscription access to this journal. Equations are available to subscribers only. You may subscribe either as an Optica member, or as an authorized user of your institution.

Contact your librarian or system administrator
or
Login to access Optica Member Subscription

Polarization State	Fourier Fit	Hippopede Method
Horizontal	$S_{0} = 1.00 \pm 0.02$	$1.00 \pm 0.00$
Horizontal	$S_{1} = 0.97 \pm 0.02$	$1.00 \pm 0.00$
	$S_{2} = 0.00 \pm 0.02$	$0.00 \pm 0.02$
Vertical	$S_{0} = 1.00 \pm 0.02$	$1.00 \pm 0.00$
Vertical	$S_{1} = - 0.97 \pm 0.02$	$- 0.99 \pm 0.002$
	$S_{2} = 0.10 \pm 0.02$	$0.07 \pm 0.02$
Anti-diagonal	$S_{0} = 1.00 \pm 0.02$	$1.00 \pm 0.00$
Anti-diagonal	$S_{1} = 0.00 \pm 0.02$	$0.00 \pm 0.02$
	$S_{2} = - 0.98 \pm 0.02$	$- 1.00 \pm 0.00$
Diagonal	$S_{0} = 1.00 \pm 0.04$	$1.00 \pm 0.00$
Diagonal	$S_{1} = 0.04 \pm 0.04$	$0.04 \pm 0.02$
	$S_{2} = 0.94 \pm 0.04$	$0.99 \pm 0.00$
Left circular	$S_{0} = 1.00 \pm 0.01$	$1.00 \pm 0.00$
Left circular	$S_{1} = 0.07 \pm 0.01$	$0.00 \pm 0.00$
	$S_{2} = - 0.01 \pm 0.01$	$0.05 \pm 0.02$
Right circular	$S_{0} = 1.00 \pm 0.02$	$1.00 \pm 0.00$
Right circular	$S_{1} = - 0.02 \pm 0.02$	$0.00 \pm 0.00$
	$S_{2} = 0.04 \pm 0.02$	$0.02 \pm 0.02$
Arbitrary	$S_{0} = 1.00 \pm 0.04$	$1.00 \pm 0.00$
Arbitrary	$S_{1} = 0.48 \pm 0.09$	$0.34 \pm 0.01$
	$S_{2} = 0.47 \pm 0.09$	$0.47 \pm 0.01$

Polarization State	One-Shot Measurement	Quantum Hippopede	Classical Hippopede
Horizontal	$S_{0} = 1.00 \pm 0.00$	$1.00 \pm 0.04$	$1.00 \pm 0.04$
	$S_{1} = 0.94 \pm 0.00$	$0.95 \pm 0.04$	$0.92 \pm 0.04$
	$S_{2} = 0.04 \pm 0.00$	$- 0.16 \pm 0.04$	$- 0.15 \pm 0.04$
Vertical	$S_{0} = 1.00 \pm 0.00$	$1.00 \pm 0.02$	$1.00 \pm 0.02$
	$S_{1} = - 0.86 \pm 0.00$	$- 0.98 \pm 0.02$	$- 0.92 \pm 0.02$
	$S_{2} = 0.20 \pm 0.00$	$- 0.08 \pm 0.02$	$- 0.09 \pm 0.02$
Anti-diagonal	$S_{0} = 1.00 \pm 0.02$	$1.00 \pm 0.00$	$1.00 \pm 0.03$
	$S_{1} = - 0.03 \pm 0.00$	$0.11 \pm 0.02$	$0.13 \pm 0.03$
	$S_{2} = - 1.00 \pm 0.00$	$- 0.97 \pm 0.02$	$- 0.93 \pm 0.03$
Diagonal	$S_{0} = 1.00 \pm 0.00$	$1.00 \pm 0.00$	$1.00 \pm 0.00$
	$S_{1} = 0.10 \pm 0.00$	$0.04 \pm 0.02$	$0.04 \pm 0.02$
	$S_{2} = 0.98 \pm 0.00$	$1.00 \pm 0.00$	$1.00 \pm 0.00$
Left circular	$S_{0} = 1.00 \pm 0.00$	$1.00 \pm 0.04$	$1.00 \pm 0.02$
Left circular	$S_{1} = - 0.12 \pm 0.00$	$- 0.04 \pm 0.04$	$0.06 \pm 0.02$
	$S_{2} = 0.03 \pm 0.01$	$- 0.02 \pm 0.04$	$- 0.03 \pm 0.02$
Right circular	$S_{0} = 1.00 \pm 0.00$	$1.00 \pm 0.04$	$1.00 \pm 0.02$
Right circular	$S_{1} = 0.07 \pm 0.00$	$- 0.07 \pm 0.04$	$0.06 \pm 0.02$
	$S_{2} = 0.09 \pm 0.00$	$- 0.04 \pm 0.06$	$- 0.04 \pm 0.02$
Arbitrary	$S_{0} = 1.00 \pm 0.00$	$1.00 \pm 0.04$	$1.00 \pm 0.04$
	$S_{1} = 0.11 \pm 0.00$	$0.33 \pm 0.16$	$0.43 \pm 0.17$
	$S_{2} = 0.48 \pm 0.00$	$0.52 \pm 0.16$	$0.44 \pm 0.17$

Polarization State	Fourier Fit	Hippopede Method
Horizontal	$S_{0} = 1.00 \pm 0.02$	$1.00 \pm 0.00$
Horizontal	$S_{1} = 0.97 \pm 0.02$	$1.00 \pm 0.00$
	$S_{2} = 0.00 \pm 0.02$	$0.00 \pm 0.02$
Vertical	$S_{0} = 1.00 \pm 0.02$	$1.00 \pm 0.00$
Vertical	$S_{1} = - 0.97 \pm 0.02$	$- 0.99 \pm 0.002$
	$S_{2} = 0.10 \pm 0.02$	$0.07 \pm 0.02$
Anti-diagonal	$S_{0} = 1.00 \pm 0.02$	$1.00 \pm 0.00$
Anti-diagonal	$S_{1} = 0.00 \pm 0.02$	$0.00 \pm 0.02$
	$S_{2} = - 0.98 \pm 0.02$	$- 1.00 \pm 0.00$
Diagonal	$S_{0} = 1.00 \pm 0.04$	$1.00 \pm 0.00$
Diagonal	$S_{1} = 0.04 \pm 0.04$	$0.04 \pm 0.02$
	$S_{2} = 0.94 \pm 0.04$	$0.99 \pm 0.00$
Left circular	$S_{0} = 1.00 \pm 0.01$	$1.00 \pm 0.00$
Left circular	$S_{1} = 0.07 \pm 0.01$	$0.00 \pm 0.00$
	$S_{2} = - 0.01 \pm 0.01$	$0.05 \pm 0.02$
Right circular	$S_{0} = 1.00 \pm 0.02$	$1.00 \pm 0.00$
Right circular	$S_{1} = - 0.02 \pm 0.02$	$0.00 \pm 0.00$
	$S_{2} = 0.04 \pm 0.02$	$0.02 \pm 0.02$
Arbitrary	$S_{0} = 1.00 \pm 0.04$	$1.00 \pm 0.00$
Arbitrary	$S_{1} = 0.48 \pm 0.09$	$0.34 \pm 0.01$
	$S_{2} = 0.47 \pm 0.09$	$0.47 \pm 0.01$

Polarization State	One-Shot Measurement	Quantum Hippopede	Classical Hippopede
Horizontal	$S_{0} = 1.00 \pm 0.00$	$1.00 \pm 0.04$	$1.00 \pm 0.04$
	$S_{1} = 0.94 \pm 0.00$	$0.95 \pm 0.04$	$0.92 \pm 0.04$
	$S_{2} = 0.04 \pm 0.00$	$- 0.16 \pm 0.04$	$- 0.15 \pm 0.04$
Vertical	$S_{0} = 1.00 \pm 0.00$	$1.00 \pm 0.02$	$1.00 \pm 0.02$
	$S_{1} = - 0.86 \pm 0.00$	$- 0.98 \pm 0.02$	$- 0.92 \pm 0.02$
	$S_{2} = 0.20 \pm 0.00$	$- 0.08 \pm 0.02$	$- 0.09 \pm 0.02$
Anti-diagonal	$S_{0} = 1.00 \pm 0.02$	$1.00 \pm 0.00$	$1.00 \pm 0.03$
	$S_{1} = - 0.03 \pm 0.00$	$0.11 \pm 0.02$	$0.13 \pm 0.03$
	$S_{2} = - 1.00 \pm 0.00$	$- 0.97 \pm 0.02$	$- 0.93 \pm 0.03$
Diagonal	$S_{0} = 1.00 \pm 0.00$	$1.00 \pm 0.00$	$1.00 \pm 0.00$
	$S_{1} = 0.10 \pm 0.00$	$0.04 \pm 0.02$	$0.04 \pm 0.02$
	$S_{2} = 0.98 \pm 0.00$	$1.00 \pm 0.00$	$1.00 \pm 0.00$
Left circular	$S_{0} = 1.00 \pm 0.00$	$1.00 \pm 0.04$	$1.00 \pm 0.02$
Left circular	$S_{1} = - 0.12 \pm 0.00$	$- 0.04 \pm 0.04$	$0.06 \pm 0.02$
	$S_{2} = 0.03 \pm 0.01$	$- 0.02 \pm 0.04$	$- 0.03 \pm 0.02$
Right circular	$S_{0} = 1.00 \pm 0.00$	$1.00 \pm 0.04$	$1.00 \pm 0.02$
Right circular	$S_{1} = 0.07 \pm 0.00$	$- 0.07 \pm 0.04$	$0.06 \pm 0.02$
	$S_{2} = 0.09 \pm 0.00$	$- 0.04 \pm 0.06$	$- 0.04 \pm 0.02$
Arbitrary	$S_{0} = 1.00 \pm 0.00$	$1.00 \pm 0.04$	$1.00 \pm 0.04$
	$S_{1} = 0.11 \pm 0.00$	$0.33 \pm 0.16$	$0.43 \pm 0.17$
	$S_{2} = 0.48 \pm 0.00$	$0.52 \pm 0.16$	$0.44 \pm 0.17$

Abstract

Cited By

Figures (6)

Tables (2)

Equations (29)

Applied Optics